FE Civil · Chapter 12 · 12–18 exam questions

FE Civil Structural Engineering

This chapter tests structural analysis, load combinations, and the design of reinforced concrete and steel members.

What the FE tests in Structural Engineering

Analysis & Load Combinations

As a civil engineer, you solve member forces in trusses (method of joints and sections), compute deflections of determinate beams, trusses, and frames (unit-load / virtual work), classify structures as determinate or indeterminate (and handle a single redundant), build influence lines for moving loads, and apply ASCE 7 load combinations to size every member.

Reinforced Concrete Design

As a civil engineer, you design reinforced concrete beams, columns, slabs, and footings on most building and infrastructure projects. You compute required flexural reinforcement using the Whitney stress block, check shear capacity and add stirrups when needed, and verify that columns can carry the factored axial load.

Steel Design

As a civil engineer, steel design means selecting W-shapes from the AISC manual to resist the required loads. You check beams for flexure and lateral-torsional buckling, columns for axial capacity and slenderness, and tension members for yielding and rupture on the net section.

Key Structural Engineering formulas

  • r+m2jr + m \geq 2j
    Truss Stability/DeterminacyFE Handbook p. 271
  • δ=nNLAE,δ=mMEIdx\delta = \sum \frac{nNL}{AE}, \quad \delta = \int \frac{mM}{EI}\,dx
    Deflection (Unit-Load Method)FE Handbook p. 271
  • δ=PL348EI,5wL4384EI,PL33EI\delta = \frac{PL^3}{48EI}, \quad \frac{5wL^4}{384EI}, \quad \frac{PL^3}{3EI}
    Standard Beam DeflectionsFE Handbook p. 271
  • Rprop=3wL8,MFE=wL212R_{\text{prop}} = \frac{3wL}{8}, \quad M_{FE} = \frac{wL^2}{12}
    Propped Cantilever / Fixed-End MomentFE Handbook p. 271
  • 1.2D+1.6L1.2D + 1.6L
    LRFD Basic CombinationFE Handbook p. 272
  • Mn=Asfy(da/2)M_n = A_s f_y (d - a/2)
    RC Beam Moment CapacityFE Handbook p. 275
  • a=Asfy0.85fcba = \frac{A_s f_y}{0.85 f_c' b}
    Whitney Stress Block DepthFE Handbook p. 275
  • Vc=2λfcbwdV_c = 2\lambda\sqrt{f_c'} \cdot b_w \cdot d
    Concrete Shear CapacityFE Handbook p. 276
  • ϕPn(max)=0.80ϕ[0.85fc(AgAst)+fyAst]\phi P_{n(\max)} = 0.80\phi[0.85 f_c'(A_g - A_{st}) + f_y A_{st}]
    RC Column Max CapacityFE Handbook p. 278
  • Mp=FyZxM_p = F_y Z_x
    Steel Plastic MomentFE Handbook p. 281
  • ϕcPn=ϕcFcrAg\phi_c P_n = \phi_c F_{cr} A_g
    Steel Column StrengthFE Handbook p. 288
  • Pn=FyAg (yield),Pn=FuAe (rupture)P_n = F_y A_g \text{ (yield)}, \quad P_n = F_u A_e \text{ (rupture)}
    Steel Tension MembersFE Handbook p. 283

Sample Structural Engineering problems

Q1. A planar truss has 9 members, 6 joints, and 3 external reactions. Is the truss statically determinate?

Answer: Yes, it is statically determinate

Explain it simply

For a truss: m+r=2jm + r = 2j for determinate. Here: 9+3=129 + 3 = 12, and 2(6)=122(6) = 12. Since m+r=2jm + r = 2j, the truss is statically determinate. If m+r>2jm + r > 2j, it is indeterminate. If m+r<2jm + r < 2j, it is unstable.

Q2. A structure is externally stable but has m+r>2jm + r > 2j. This means the structure is:

Answer: Statically indeterminate

Explain it simply

When m+r>2jm + r > 2j, there are more unknowns than equilibrium equations. The structure has redundant members or supports -- it is statically indeterminate. You would need compatibility equations in addition to equilibrium to solve it. The degree of indeterminacy is (m+r)2j(m + r) - 2j.

These are 2 of 1,126 problems across all 15 chapters. The full bank, lessons, mastery tracking, and timed exam simulation live inside the app.

Common Structural Engineering mistakes on the FE

  • Using ASD factors when the problem asks for LRFD, or vice versa — read the load combination method carefully.
  • Forgetting that d (effective depth) is NOT the same as the total beam height h — d = h minus cover minus bar offset.
  • Using Zx (plastic section modulus) vs. Sx (elastic section modulus) — LRFD steel uses Zx, ASD uses Sx.
  • Not checking all ASCE 7 load combinations — the controlling case often is not the one with the largest single load.
  • Picking the LARGER capacity as the controlling limit state — the SMALLER design strength governs because it represents the weaker failure mode.
  • Point-load deflection uses L³ (PL³/48EI); distributed-load deflection uses L⁴ (5wL⁴/384EI) — mixing the powers of L is the #1 deflection error.
  • Confusing the propped-cantilever prop reaction 3wL/8 with the simple-span reaction wL/2, and the fixed-end moment wL²/12 with the simple-span moment wL²/8.

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